NDA Maths · Differentiation

Differentiability — When the Derivative Exists

A function is differentiable at a point only if its left-hand and right-hand derivatives both exist and are equal — corners, jumps, and steps are where this fails.

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Why this matters

NDA loves to test the gap between continuity and differentiability. The modulus, piecewise, and greatest-integer functions are the standard counter-examples, and parameter problems ('find a, b so f is differentiable') hinge entirely on matching one-sided derivatives.

Concept 1 of 6

Differentiable implies continuous (not the converse)

Intuition

Differentiability is the stronger condition. If a function has a derivative at a point, it must be continuous there — but a continuous function can still fail to be differentiable (a sharp corner). So 'differentiable

\Rightarrow

continuous' is a one-way street.

Definition

If $f$ is differentiable at $x=c$ , then $f$ is continuous at $c$ .
The converse is false: $|x|$ is continuous at $0$ but not differentiable there.
Contrapositive (useful): if $f$ is discontinuous at $c$ , it cannot be differentiable at $c$ .

Worked example

A function has a jump discontinuity at

x=2

. Can it be differentiable at

x=2

Differentiability requires continuity first (differentiable $\Rightarrow$ continuous).
The function is discontinuous at $x=2$ , so the necessary condition already fails.

Answer:No — a discontinuous point can never be a differentiable point.

Practice this conceptself-check · 4 quick reps

Try it yourself

True or false: every continuous function is differentiable.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Differentiable at $c$ $\Rightarrow$ ?
2.
Continuous at $c$ $\Rightarrow$ differentiable at $c$ ?
3.
If $f$ is discontinuous at $c$ , is it differentiable there?
4.
Canonical continuous-but-not-differentiable function?

The implication only runs one way

'Differentiable

\Rightarrow

continuous' is true; 'continuous

\Rightarrow

differentiable' is FALSE. NDA statement-questions plant the reversed (false) version to catch you.

Drill 1 more on differentiable implies continuous (not the converse)

Concept 2 of 6

The left-hand = right-hand derivative test

Intuition

At a join point of a piecewise function, compute the derivative from the left and from the right separately. The function is differentiable there only if both exist and are equal. This is the workhorse for every piecewise differentiability question.

Definition

$f$ is differentiable at $x=c$ iff the one-sided derivatives match:

LHD $= \lim_{h\to 0^-}\dfrac{f(c+h)-f(c)}{h}$
RHD $= \lim_{h\to 0^+}\dfrac{f(c+h)-f(c)}{h}$

Differentiable at $c$ $\iff$ LHD $=$ RHD (and both finite). In practice: differentiate each piece, then equate the two pieces' derivatives at the join (after first checking continuity there).

One-sided derivative test

f'(c^-) = f'(c^+) \iff f \text{ differentiable at } c

Worked example

f(x)=\begin{cases} x^2, & x\le 1\\ 2x-1, & x>1\end{cases}

differentiable at

x=1

Continuity at $1$ : left value $1^2=1$ , right value $2(1)-1=1$ — continuous.
LHD: derivative of $x^2$ is $2x$ , at $x=1$ gives $2$ .
RHD: derivative of $2x-1$ is $2$ , at $x=1$ gives $2$ .
LHD $=$ RHD $=2$ .

Answer:Yes — differentiable at

x=1

(the pieces meet smoothly).

Practice this conceptself-check · 4 quick reps

Try it yourself

For

f(x)=\begin{cases} x^2, & x\le 1\\ x+1, & x>1\end{cases}

, is

f

differentiable at

x=1

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Condition for differentiability at a join?
2.
Check before computing one-sided derivatives?
3.
RHD of $2x-1$ at $x=1$ ?
4.
If LHD $=3$ , RHD $=5$ , differentiable?

From the bank · past-year question

Example 2DifferentiationMODERATE

A function is defined in

(0, \infty)

f(x) = \begin{cases}1 - x^2 & \text{for } 0 < x \leq 1 \\ \ln x & \text{for } 1 < x \leq 2 \\ \ln 2 - 1 + 0.5x & \text{for } 2 < x < \infty\end{cases}

. Which one of the following is correct in respect of the derivative of the function, i.e.,

f'(x)

[Q68 · Sep · 2017]

Continuity first, then slopes

If the pieces don't even meet (discontinuous at the join), stop — it cannot be differentiable. Only when continuous do you compare LHD and RHD.

Drill 6 more on the left-hand = right-hand derivative test

Concept 3 of 6

Modulus corners — and the x|x| trap

Intuition

A modulus makes a corner at its split point, where the slope jumps.

|x|

has slope

-1

on the left and

+1

on the right, so it is not differentiable at

0

. But not every modulus expression has a corner:

x|x|

smooths the kink out and is differentiable at

0

. Always split at the modulus's zero and test.

Definition

Split $|g(x)|$ at the points where $g(x)=0$ :

$|x|$ is continuous but not differentiable at $0$ (LHD $=-1$ , RHD $=+1$ ).
$|x-a|$ has its corner at $x=a$ .
$x|x|$ equals $-x^2$ for $x<0$ and $x^2$ for $x\ge 0$ ; both have derivative $0$ at the join, so it is differentiable at $0$ (the trap).

Worked example

f(x)=|x-3|

differentiable at

x=3

For $x>3$ : $f(x)=x-3$ , slope $+1$ . For $x<3$ : $f(x)=3-x$ , slope $-1$ .
LHD $=-1$ , RHD $=+1$ — not equal.

Answer:No — there is a corner at

x=3

(continuous, not differentiable).

Practice this conceptself-check · 4 quick reps

Try it yourself

f(x)=x|x|

differentiable at

x=0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Is $|x|$ differentiable at $0$ ?
2.
Where is $|x-5|$ non-differentiable?
3.
Is $x|x|$ differentiable at $0$ ?
4.
Is $|x|$ continuous at $0$ ?

From the bank · past-year question

Example 3DifferentiationMODERATE

For the following three (03) items: Consider the function

f(x)=x|x|

Consider the following statements: I. The function is increasing in the interval

(-\infty,\infty)

. II. The function is differentiable at

x=0

. Which of the statements given above is/are correct?

[Q75 · Sep · 2025]

Not every |·| means non-differentiable

|x|

has a corner, but

x|x|

x^2|x|

, and

e^{|x|}\cdot(\text{smoothing})

can be differentiable at the split. Always rewrite piecewise and compare one-sided derivatives — don't assume the modulus kills differentiability.

Drill 3 more on modulus corners — and the x|x| trap

Concept 4 of 6

Greatest-integer and step functions

Intuition

The greatest-integer function

\lfloor x\rfloor

is a staircase — flat between integers, with a jump at each integer. On any flat step it is constant, so its derivative is

0

; at each integer it is discontinuous, hence not differentiable.

Definition

Between consecutive integers, $\lfloor x\rfloor$ is constant $\Rightarrow$ derivative $=0$ there.
At every integer it jumps (discontinuous) $\Rightarrow$ not differentiable at integers.
A function built from $\lfloor x\rfloor$ inherits these jumps unless they cancel; check each integer in the domain.

Worked example

What is

\dfrac{d}{dx}\lfloor x\rfloor

for

2<x<3

On $(2,3)$ , $\lfloor x\rfloor = 2$ , a constant.
The derivative of a constant is $0$ .

Answer:

0

(and the derivative does not exist at the integers

2, 3

Practice this conceptself-check · 4 quick reps

Try it yourself

y=\lfloor x+1\rfloor

differentiable on the open interval

(-4,-3)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\frac{d}{dx}\lfloor x\rfloor$ on $(1,2)$ ?
2.
Is $\lfloor x\rfloor$ differentiable at $x=3$ ?
3.
Is $\lfloor x\rfloor$ continuous on $(0,1)$ ?
4.
Derivative of any constant function?

From the bank · past-year question

Example 4DifferentiationEASY

Let

y=[x+1]

-4<x<-3

where

[.]

is the greatest integer function. What is the derivative of

y

with respect to

x

x=-3.5

[Q71 · Apr · 2022]

Concept 5 of 6

Finding parameters so f is differentiable

Intuition

When a piecewise function carries unknown constants and is declared differentiable at the join, you get two equations: continuity (the values match) and differentiability (the one-sided derivatives match). Solve them together for the unknowns.

Definition

For a piecewise $f$ differentiable at $x=c$ : 1. Continuity: set the two pieces' values equal at $c$ . 2. Differentiability: set the two pieces' derivatives equal at $c$ . Two equations in the unknown constants — solve simultaneously.

Worked example

Find

a, b

so that

f(x)=\begin{cases} x^2, & x\le 1\\ ax+b, & x>1\end{cases}

is differentiable at

x=1

Continuity at $1$ : $1 = a+b$ .
Differentiability at $1$ : LHD $=2x\big|_{1}=2$ , RHD $=a$ , so $a=2$ .
Then $b = 1-a = -1$ .

Answer:

a=2,\ b=-1

Practice this conceptself-check · 4 quick reps

Try it yourself

f(x)=\begin{cases} ax+1, & x\le 0\\ \sin x + b, & x>0\end{cases}

is differentiable at

0

, find

a, b

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
How many equations from 'differentiable at a join'?
2.
Continuity equation gives?
3.
Differentiability equation gives?
4.
For $x^2$ vs $ax+b$ at $x=1$ : the slope equation?

From the bank · past-year question

Example 5DifferentiationHARD

Let

f(x)=\frac{ax}{x+1}+b,\ x<1

and

\sqrt{x-1},\ 1\leq x\leq2

f(x)

is differentiable at

x=1

, then what is the value of

(a+b)

[Q63 · Sep · 2023]

Concept 6 of 6

Derivative at an awkward point via the limit definition

Intuition

When a point is special — a modulus, a

\ln|x|

, or a value patched in by hand (

f(0)=0

) — the rule-based derivative may not apply directly. Fall back to the definition:

f'(c)=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}

and evaluate the limit.

Definition

At a point where the usual rules are unsafe, compute $f'(c)=\lim_{h\to 0}\dfrac{f(c+h)-f(c)}{h}$ directly. If the limit exists (and is the same from both sides), that value is the derivative; if it doesn't, $f$ is not differentiable at $c$ .

Worked example

For

f(x)=x^2\ln|x|

with

f(0)=0

, find

f'(0)

Use the definition: $f'(0)=\lim_{h\to 0}\dfrac{h^2\ln|h| - 0}{h}=\lim_{h\to 0} h\ln|h|$ .
As $h\to 0$ , $h\ln|h|\to 0$ (the linear factor beats the log).

Answer:

f'(0)=0

Practice this conceptself-check · 4 quick reps

Try it yourself

For

f(x)=x|x|

, find

f'(0)

from the definition.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Definition of $f'(c)$ ?
2.
$\lim_{h\to 0} h\ln|h|$ ?
3.
When to fall back to the definition?
4.
If the two-sided limit fails, then?

From the bank · past-year question

Example 6DifferentiationMODERATE

For

f(x)=x^2\ln|x|

(

x\ne0

f(0)=0

, what is

f'(0)

[Q74 · Apr · 2018]

Drill 1 more on derivative at an awkward point via the limit definition

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

The left-hand = right-hand derivative test
One-sided derivative test
$f'(c^-) = f'(c^+) \iff f \text{ differentiable at } c$

Watch out for (3)

The implication only runs one way
→ Differentiable implies continuous (not the converse)
Continuity first, then slopes
→ The left-hand = right-hand derivative test
Not every |·| means non-differentiable
→ Modulus corners — and the x|x| trap

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1DifferentiationMODERATE

Suppose the function

f(x)=x^n

n\neq0

is differentiable for all

x

. Then

n

can be any element of the interval

[Q99 · Apr · 2017]

Example 2DifferentiationEASY

Consider the following for the items that follow: Let

f(x)=|\ln x|,\, x\neq1

What is the derivative of

f(x)

x=0.5

[Q86 · Apr · 2023]

Example 3DifferentiationMODERATE

What is

\frac{d\sqrt{1-\sin 2x}}{dx}

equal to, where

\frac{\pi}{4}<x<\frac{\pi}{2}

[Q96 · Sep · 2018]

Example 4DifferentiationHARD

The set of all points, where the function

f(x) = \sqrt{1 - e^{-x^2}}

is differentiable, is

[Q83 · Sep · 2017]

Example 5DifferentiationEASY

For the following two (02) items: Let the function

f(x)=|x-3|+|x-4|

be defined on the interval

[0,5]

What is

\dfrac{dy}{dx}

x=3.5

equal to?

[Q87 · Sep · 2025]

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